Lemma 1

— When $\mathscr{L}_0$ and $\mathscr{K}(T)$ are equipped with the topology induced by that of $\mathscr{L}^1((\mu_\iota)_{\iota \in I})$: a) each element of $\mathscr{L}_0$ is in the closure of some bounded subset of $\mathscr{K}(T)$; b) each element of $\mathscr{L}^1((\mu_\iota)_{\iota \in I})$ is in the closure of some bounded subset of $\mathscr{L}_0$.

To prove a), we may restrict ourselves to the case of an element of the form $f = g \varphi_K$ ($g \in \mathscr{C}(T)$), K compact in T). It is immediate (by virtue of Urysohn’s theorem) that $f$ is in the closure of the set B of functions of the form $gh$, where h describes the set of continuous mappings of T into $[0,1]$ that are equal to 1 on K and to 0 on the complement of a fixed compact neighborhood H of K. Moreover, the set B is bounded, because $\mu_\iota(|gh|) \leq \mu_\iota(|f\varphi_H|)$ for every function h having the preceding properties.

Let us now prove b); we may restrict ourselves to the case of an element $f \geq 0$ of $\mathscr{L}^1((\mu_\iota)_\iota \in I)$. For every $\iota \in I$ and every $\varepsilon > 0$, there exists a compact subset $K(\iota, \varepsilon)$ of T such that the restriction of $f$ to $K(\iota, \varepsilon)$ is continuous and $|\mu_\iota(|f - f\varphi_{K(\iota, \varepsilon)}|)| \leq \varepsilon$. It is clear that $f$ is in the closure of the set C of $f\varphi_{K(\iota, \varepsilon)}$ (where $\iota \in I,\ \varepsilon > 0$). By virtue of Urysohn’s theorem, the set C in contained in $\mathscr{L}_0$; moreover, it is bounded, because $\mu_\kappa(f\varphi_{K(\iota, \varepsilon)}) \leq \mu_\kappa(f)$ for all $\iota \in I,\ \kappa \in I$ and $\varepsilon > 0$.

Let us now prove Prop. 6: for every function $g \in \mathscr{K}(T)$ and every continuous seminorm q on F, $q\left( \int g\,dm \right) \leq \int |g|\,d(q(m))$ (Prop. 5), which implies that the mapping $g \mapsto \int g\,dm$ of $\mathscr{K}(T)$ into F is continuous when $\mathscr{K}(T)$ is equipped with the topology induced by that of $\mathscr{L}^1((q(m))_{q \in Q})$ (Q the set of continuous semi-norms on F). Consequently, by the preceding lemma and Prop. 10 of TVS, III, §1, No. 6, this mapping may be extended by continuity, first to a continuous linear mapping $v_0$ of $\mathscr{L}_0$ into F, then to a continuous linear mapping $v$ of $\mathscr{L}^1((q(m))_{q \in Q})$ into F. Moreover, for every $z' \in F'$ the relation $\langle z', v(f) \rangle = \int f\,d(z' \circ m)$ holds, by the definition of $v$, for every $f \in \mathscr{K}(T)$; since $|z' \circ m| \leq q(m)$ for $q(z) = |\langle z', z \rangle|$, the mapping $f \mapsto \int f\,d(z' \circ m)$ is continuous on $\mathscr{L}^1((q(m))_{q \in Q})$, therefore again by continuity, the relation $\langle z', v(f) = \int f\,d(z' \circ m)$ holds for every function $f \in \mathscr{L}^1((q(m))_{q \in Q})$. It follows that $v(f) = \int f\,dm$, which completes the proof.